# For a differential equation, prove that there exists an integrating factor dependent only on $x$

Consider the differential equation $$M(x, y)dx + N(x,y)dy = 0$$. Prove that there exists an integrating factor that is dependent on $$x$$ if and only if $$\frac {M_y - N_x}{N} = f(x)$$, a function of $$x$$ only.

I need to prove that in such a case, the integrating factor is: $$I(x) = e^ { \int {f(x)dx}}.$$

I tried to approach this problem by rewriting $$M(x, y)dx + N(x,y)dy = 0$$ as $$M(x, y) + N(x,y) \frac {dy}{dx} = 0.$$ Assuming the differential equation is exact, there exists a potential function $$φ$$ such that $$\frac {∂φ}{∂x} + \frac{∂φ}{∂y} \frac{dy}{dx} = 0.$$

But this implies that $$\frac {∂φ}{∂x} = 0$$. So $$φ(x, y)$$ is a function of $$y$$ only. So we can deduce that $$φ(x, y)$$ is a function of $$x$$ only and $$φ(x, y) = c$$, where $$c$$ is a constant.

This is where I'm stuck and I'm not sure how to proceed with this proof. Any guidance is greatly appreciated!

• Is it given that the differential equation is exact? Commented Feb 1, 2021 at 6:59
• @P. J. No, I just assumed that fact because I had no idea how to approach the problem without it.
– user815455
Commented Feb 1, 2021 at 7:04
• I don't think one should assume that it is exact - if it is then $M_y = N_x \implies M_y - N_x =$ and one cannot proceed further with the proof Commented Feb 1, 2021 at 7:08
• Taking your comment into consideration, I'm still not sure how to proceed. Could you provide some guidance?
– user815455
Commented Feb 1, 2021 at 10:00

Question: Consider the differential equation $$M(x, y)dx + N(x,y)dy = 0$$. Prove that there exists an integrating factor that is dependent on $$x$$ if and only if $$\frac {1}{N}(M_y - N_x) = f(x)$$, a function of $$x$$ only.
Proof: Here the given general first order differential equation is $$M(x, y)dx + N(x,y)dy = 0\tag1$$ We shall suppose that there exists an integrating factor that depends only on $$~x~$$: $$~\mu=\mu(x)~$$
Now if $$~\mu~$$ is an integrating factor, then $$~M(x, y)\mu(x)dx + N(x,y)\mu(x)dy~$$ must be exact. i.e., $$\dfrac{\partial}{\partial y}\left[M(x, y)\mu(x)\right]=\dfrac{\partial}{\partial x}\left[N(x, y)\mu(x)\right]$$ $$\implies\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$$ $$\implies \dfrac{\partial\mu}{\partial x}=\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)\mu\tag2$$ Now if $$~\mu~$$ depends only on $$~x~$$ and not on $$~y~$$, then necessarily $$~\mu~$$ depends only on $$~x~.$$ Thus the RHS of the above equation must be a function of $$~x~$$ only. i.e., $$\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)=f(x)$$ Thus the equation $$(2)$$ becomes., $$\dfrac{\partial\mu}{\partial x}-f(x)\mu=0$$ which is a first order linear differential equation and can be solved and the solution is $$\mu(x)=A \exp\left[\int f(x) dx\right]=A \exp\left[\int \dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right) dx\right]$$which gives us an integrating factor for the equation $$(1)$$ in the form $$~\frac 1{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)~$$ depends on $$~x~$$ only.
Conversely, let for the equation $$(1)$$, $$\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)=f(x)\implies Nf(x)=\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\tag3$$ Multiplying both side of $$(1)$$ by $$~e^{\int f(x)dx}~$$, we have $$~M_1~dx+N_1~dy=0~$$ where $$~M_1=M~e^{\int f(x)dx}~$$ and $$~N_1=N~e^{\int f(x)dx}~.$$ Now $$\dfrac{\partial M_1}{\partial y}=\dfrac{\partial M}{\partial y}e^{\int f(x)dx}\tag4~~~~~~~~~~\text{and}$$ $$\dfrac{\partial N_1}{\partial x}=\dfrac{\partial N}{\partial x}e^{\int f(x)dx}+N~e^{\int f(x)dx}~f(x)=e^{\int f(x)dx}\left\{\dfrac{\partial N}{\partial x}+Nf(x)\right\}$$ $$\implies \dfrac{\partial N_1}{\partial x}=e^{\int f(x)dx}\left\{\dfrac{\partial N}{\partial x}+\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right\}~,~~~\text{from (3)}$$ $$\implies \dfrac{\partial N_1}{\partial x}=e^{\int f(x)dx}~\dfrac{\partial M}{\partial y}\tag5$$ From equation $$(4)$$ and $$(5)$$, it is clear that $$\dfrac{\partial M_1}{\partial y}=\dfrac{\partial N_1}{\partial x}$$which shows that $$~M_1~dx+N_1~dy=0~$$ is an exact differential equation and hence $$~e^{\int f(x)dx}~$$ is it's integrating factor that depends only on $$~x~$$.